Problem: What is the value of the following logarithm? $\log_{7} \left(\dfrac{1}{343}\right)$
Answer: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $7^{y} = \dfrac{1}{343}$ In this case, $7^{-3} = \dfrac{1}{343}$, so $\log_{7} \left(\dfrac{1}{343}\right) = -3$.